A134434 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k even entries that are followed by a smaller entry (n>=0, k>=0).

1, 1, 1, 1, 4, 2, 4, 16, 4, 36, 72, 12, 36, 324, 324, 36, 576, 2592, 1728, 144, 576, 9216, 20736, 9216, 576, 14400, 115200, 172800, 57600, 2880, 14400, 360000, 1440000, 1440000, 360000, 14400, 518400, 6480000, 17280000, 12960000, 2592000, 86400

Offset: 0

Author

Emeric Deutsch, Nov 22 2007

Keywords

• nonn
• tabf

Comments

Row n has 1+floor(n/2) entries. T(2n-1,0) = T(2n,0) = T(2n,n) = (n!)^2 = A001044(n).

This descent statistic is equidistributed on the symmetric group S_n with a multiplicative 2-excedance statistic - see A136715 for details. - Peter Bala, Jan 23 2008

Examples

			T(4,2) = 4 because we have 2143, 4213, 3421 and 4321.
Triangle starts:
   1;
   1;
   1,   1;
   4,   2;
   4,  16,   4;
  36,  72,  12;
  36, 324, 324, 36;
  ...
		

Links

• S. Kitaev and J. Remmel, Classifying descents according to parity, Annals of Combinatorics, 11, 2007, 173-193.

Crossrefs

Bisection of column k=0 gives A001044 (even part).

Row sums give A000142.

Cf. A134435, A136715.

Programs

Maple

R[0]:=1:R[1]:=1: R[2]:=1+t: for n to 5 do R[2*n+1]:=sort(expand((1-t)* (diff(R[2*n], t))+(2*n+1)*R[2*n])): R[2*n+2]:=sort(expand(t*(1-t)*(diff(R[2*n+1], t))+(1+(2*n+1)*t)*R[2*n+1])) end do: for n from 0 to 11 do seq(coeff(R[n], t, j), j=0..floor((1/2)*n)); end do; # yields sequence in triangular form

Mathematica

T[n_,k_]:=If[EvenQ[n],Floor[(n/2)!Binomial[n/2,k]]^2, Floor[((n+1)/2)!Binomial[(n-1)/2,k]]^2/(k+1)]; Table[T[n,k],{n,11},{k,0,Floor[n/2]}]//Flatten (* Stefano Spezia, Jul 12 2024 *)

Formula

T(2n,k) = [n!*C(n,k)]^2; T(2n+1,k) = [(n+1)!*C(n,k)]^2/(k+1). See the Kitaev & Remmel reference for recurrence relations (Sec. 3).

Extensions

T(0,0)=1 prepended by Alois P. Heinz, Jul 12 2024